Does the function $U=\frac{kx}{(x+x_{0})^2}$ reduce to a simple harmonic potential energy function for small oscillations around $x=x_0$?

Question

The question is from a physics problems book,

A particle of mass $m$ moves in a potential energy function given by $$U=\frac{kx}{(x+x_{0})^2}$$ where $x$ denotes the position and $x_0$ is a constant.

1. What is the stable equilibrium position of the particle?
2. What will be the time period of small oscillations about this mean position?

The first part is easy. The answer to the first part is

$$x=x_0 $$

for stable equilibrium.

The solution is by first differentiating the potential energy function and setting it to zero. We obtain two answers $x_0$ and $-x_0$.

$-x_0$ is rejected as a solution since it is in unstable equilibrium.

The second part of the question is baffling me.

I am unable to prove that the particle will perform SHM for small oscillations around $x=x_0$ even by using approximations.

To prove that a particle performs SHM we can proceed by two cases, both conditions are sufficient by them self.

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Potential Energy $U$ should be of the form $U=Ax^2+Bx$ where $A>0$

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Force on the particle should be of the form $F=Px+Q$ where $P<0$

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Note that force and potential energy are related by $F=-\frac{dU}{dx}$

So what is my mathematical question?

Can you show that the expression $U=\frac{kx}{(x+x_{0})^2}$ reduces to the two cases above by using the fact that the particle is performing small oscillations around the point $x=x_0$?

I think that some mathematical approximations will be needed to simplify the expression.

I graphed the function on Desmos and what I saw was this. By adjusting the sliders you can see that the graph has a infinite well or infinite peak. I don't think any particle can perform SHM in such a potential energy distribution. Usually for SHM the potential energy distribution is of a form where there is a finite well in the graph like this.

The answer given in the book is

$$T=2\pi\sqrt{\frac{2mx_{0}^3}{k}}$$

Answer 1

$$F=\frac{d}{dx}\Big(\frac{kx}{(x+x_0)^2}\Big)=\frac{-kx+kx_0}{(x+x_0)^3}=0$$ .: $$x=x_0$$ So, $$a=\frac{-kx+kx_0}{m(x+x_0)^3}$$ At $x=x_0$$$x+x_0\approx 2x_0$$and$$x =x_0+\delta x$$.:$$a=-\frac{k}{m(2x_0)^3}\delta x$$